Quantitative Titration

  • Thread starter jessica.so
  • Start date
  • #1
20
0
If you were given a sample unknown solid acid that had a mass of 0.7 g and it was being dissolved in 50 mL of water and was being titrated with NaOH, is it possible to find the molar mass or the formula of the acid?

Given information
Mass of solid acid: 0.7 g
Volume of acid solution: 50 mL = 0.05 L
Volume of base used: 29.5 mL = 0.0295 L
[NaOH] = 0.095 mol/L

I was also told that this is monoprotic.
HX (aq) + NaOH (aq) --> H2O (l) + NaX (aq)
 

Answers and Replies

  • #2
16
0
Hi,
I would try to get the Acid dissociation constant with this little formula:
Ka=( [A-]*[H+])/[HA]
How many H+ were in the solution?
How many mol are 0,7 g H+ ?= c (O)
HA = c0 H+ - [H+].
Then you just have to search for the specific the ka in and you have you acid.
Greetings Firelion
 
  • #3
Borek
Mentor
28,701
3,188
No need for acid dissociation constant, in fact it is completely unrelated to problem.

If acid is monoprotic you can easily calculate number of moles from the neutralization stoichiometry. Then you know mass of the sample and number of moles in the sample - that's enough to calculate molar mass.

--
 
  • #4
16
0
oh sorry.
We leraned in our exams to calculate very different types of data with the Ka so I thought this task is similar.
 

Related Threads on Quantitative Titration

  • Last Post
Replies
9
Views
25K
Replies
3
Views
3K
Replies
1
Views
2K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
13
Views
6K
Top